Will Cooper wrote:
> Yes, there is! See the paper by M.K. Simon referenced below where he shows
> how the CDF of a Gaussian, i.e.
>
> Gaussian Q function , Q(x) =1/(2*pi)* Integral(exp(-x^2/2)) dx, x, infinity
>
> can be represented by:
>
> Q(x) = 1/pi * Integral(exp(-x^2/(2*(sin(Theta))^2)) dTheta, x, pi/2
>
> Ref: Proceedings of the IEEE, Vol 86, No9, Sep 1998 " A Unified Approach to
> the Performance Analysis of Digital Communications of Generalized Fading
> Channels", (Eqns.1 & 2, pp1863-1864).
>
> Hope this helps,
Actually, I don't see how this helps at all! Firstly, Mathematica can compute
the first integral, (which should read, if I've made the right corrections)
1/(2*Pi)*Integrate[Exp[-y^2/2], {y, x, Infinity}]
in closed form in terms of the error function.
Secondly, the alternative form as written is not equivalent to the first
integral (check this for yourself). As far as I can see, the reason that some
people might think the second form is preferable is because the domain of
integration is finite. Since Mathematica can handle infinite domains, this is
not really an issue.
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Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia
Nedlands WA 6907 mailto:paul at physics.uwa.edu.au
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God IS a weakly left-handed dice player
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